How do you prove this equation is a trigonometric identity?

Science & Mathematics by Anonymous 2018-05-18 11:06:36

Social Science

7 answers

(secx-cosx)/(cosx)=tan^2x Lmao it is hard

Anonymous

LHS= (1/cosX-cosX)/cosX= 1/cos^2(X)-1= (1-cos^2(X))/cos^2(X)= sin^2(X)/cos^2(X)= [sin(X)/cos(x)]^2= tan^2(X)= RHS
Anonymous

(secx-cosx)/(cosx) = tan^2 x (sec x/cos x) - (cos x/cos x) = tan ^2 x Let sec x = 1/cos x (1/cos x)/(cos x) - 1 = tan^2 x 1/cos^2 x - 1 = tan^2 x sec^2 x - 1 = tan^2 x Done!
Anonymous

(sec(x) - cos(x)) / cos(x) (LHS) = (sec(x)/cos(x)) - (cos(x)/cos(x)) = ((1/cos(x))/cos(x)) - 1 = (1/cos^2(x)) - 1 = sec^2(x) - 1 = tan^2(x) (RHS). Done.
Anonymous
Anonymous

(sec x - cos x ) /cos x = sec x / cos x - cos x /cos x = sec x sec x - 1 = sec^2 x - 1 = tan^2 x
Anonymous

(secx - cosx)/cosx = secx/cosx - cosx/cosx = sec^2x - 1 = tan^2x
Anonymous

1/cos^2(x) - 1 = [1-cos^2(x)]/cos^2(x) = sin^2(x)/cos^2(x) = tan^2(x)